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I want to find a semi-constructive example of a unitary commutative ring without any maximal ideals assuming that axiom of choice is incorrect and/or a model of $\sf ZF$ where we have such a concrete ring.

This question is similar to the questions Vector space bases without axiom of choice and A confusion about Axiom of Choice and existence of maximal ideals..

What I tried is to use $\mathbb{R}$ as a $\mathbb{Q}$ vector space without a basis and try to construct some chains of ideals on a related ring and try to show that a maximal ideal corresponds to a basis but didn't achieve much.

Thank you in advance!

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  • $\begingroup$ As far as I know, existence of maximal ideals in rings is properly weaker than choice. $\endgroup$
    – egreg
    Jul 7, 2019 at 13:59
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    $\begingroup$ @egreg math.stackexchange.com/questions/317028/… - they seem equal $\endgroup$
    – aeyalcinoglu
    Jul 7, 2019 at 14:24
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    $\begingroup$ I stand corrected. $\endgroup$
    – egreg
    Jul 7, 2019 at 14:28
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    $\begingroup$ Well. You can either construct special models with particular examples, and that normally involves techniques like forcing and stuff; or you can sort of repeat the proof that the existence of maximal ideals implies choice and start with a set that cannot be well-ordered as a counterexample. This set can be $\Bbb R$ in models where there is no Hamel basis, for example. $\endgroup$
    – Asaf Karagila ♦
    Jul 7, 2019 at 14:46
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    $\begingroup$ Near-duplicate: math.stackexchange.com/questions/2182014/… $\endgroup$
    – Eric Wofsey
    Jul 7, 2019 at 18:29

1 Answer 1

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Let $k$ be a field, let $I\subset k^{\mathbb{N}}$ be the ideal of sequences that are eventually zero, and let $S=k^{\mathbb{N}}/I$. Then maximal ideals in $S$ are in bijection with nonprincipal ultrafilters on $\mathbb{N}$. In particular, in any model of ZF in which there are no nonprincipal ultrafilters on $\mathbb{N}$, there will be no maximal ideals in $S$.

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